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ENGINEERING, CS 2019 HSC 21d

A pictorial sketch of a laminated timber beam used in a bus shelter is shown. This beam has a second moment area of `117 × 10^(-6)\ text{m}^(4)`. The maximum applied bending moment is 17 kNm.
 

  1. Calculate the maximum bending stress present in the beam.   (3 marks)
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`18.2\ text{MPa}`

Show Worked Solution

Second Moment of Area (`l`)`= 117 times 10^(-6)\ text{m}^(4)`

Maximum Bending Moment (`M`)`= 17 times 10^(3)\ text{N}`

Distance (Neutral axis → outside of beam) = 125 mm

`=> y = 0.125\ text{m}`

`:.\ text{Bending Stress}_text{max}` `=(My)/l`  
  `=(0.125 times 17 times 10^(3))/(117 times 10^(-6)`  
  `=18.2\ text{MPa}`  

ENGINEERING, CS 2018 HSC 23c

The chassis for a motor vehicle could be made from a solid steel bar or hollow steel tube. Both the bar and the tube have the same mass and cost per metre.
 

Justify why a hollow tube would be used in preference to a solid bar to build the vehicle chassis.   (3 marks)

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→ Hollow tube has a higher second moment of area and is therefore more rigid.

→ Because the chassis is exposed to continual flexural forces and moments when the vehicle is moving, rigidity is important.

→ Stiffness is therefore a major design requirement.

→ The hollow tube also provides damage protection for any electrics cabling.

Show Worked Solution

→ Hollow tube has a higher second moment of area and is therefore more rigid.

→ Because the chassis is exposed to continual flexural forces and moments when the vehicle is moving, rigidity is important.

→ Stiffness is therefore a major design requirement.

→ The hollow tube also provides damage protection for any electrics cabling.

♦♦ Mean mark 35%.

ENGINEERING, CS 2021 HSC 14 MC

The diagram shows a beam.

The beam's second moment of area is 76.96 × 106 mm4. It is subjected to a maximum bending moment of 250 kN m.

What is the maximum bending stress of the beam?

  1. 227.4 MPa
  2. 324.8 MPa
  3. 649.7 MPa
  4. 9745.3 MPa
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`B`

Show Worked Solution

`M=250\ text{kNm}\ = 250 xx10^(-3)\ text{Nm}`

`y=(200)/(2)\ text{mm}\ =100 xx10^(-3)\ text{m}`

`I=76.96 xx10^(6)\ text{mm}^(4)=76.96 xx10^(-6)\ text{m}^(4)`

`S_text{bending}` `=(My)/I`  
  `=(250 xx10^(-3)xx100 xx10^(-3))/(76.96 xx10^(-6))`  
  `=324.8440…\ text{MPa}`  

 
`=>B`